Chain rule special application I am given the equation 
$$ u(x,t)_t = u(x,t)_x $$
and i have to apply coordinate transformation with 
$$ x=x(\xi,\theta), \quad t=\theta $$
to get an equation of the form 
$$ \alpha u_{\theta} +\beta x_{\theta} = \gamma u_{\xi}$$
I am having trouble applying the chain rule here.
I tried:
$$ \frac{\partial u(x,t)}{\partial t} = \frac{\partial u(\xi,\theta)}{\partial \xi} \frac{\partial \xi}{\partial t} + \frac{\partial u(\xi,\theta)}{\partial \theta} \frac{\partial \theta}{\partial t} = u_{\xi}\frac{\partial \xi}{\partial t} + u_{\theta}\cdot1 $$
and 
$$
\frac{\partial u(x,t)}{\partial x} = \frac{\partial u(\xi,\theta)}{\partial \xi} \frac{\partial \xi}{\partial x} + \frac{\partial u(\xi,\theta)}{\partial \theta} \frac{\partial \theta}{\partial x} = u_{\xi}\frac{\partial \xi}{\partial x} + u_{\theta}\cdot0
$$
Thanks for the answer. 
My next problem is the Jacobian determinant because i don't know how to to interprete the coordinate transformation. It is not as usual, e.g., $(x,y) \to (u,v)$ or $(x,y) \to (r,\phi)$. The matrix is probably as follows
$$
\begin{pmatrix}
\frac{\partial x}{\partial x(\xi,\theta)}   & \frac{\partial x}{\partial \theta} \\
\frac{\partial t}{\partial x(\xi,\theta)}   &  \frac{\partial t}{\partial \theta}
\end{pmatrix}
=
\begin{pmatrix}
\frac{\partial x}{\partial x(\xi,\theta)}   & x_{\theta} \\
\frac{\partial t}{\partial x(\xi,\theta)}   &  1
\end{pmatrix}
$$ 
but how do i interprete $\frac{\partial x}{\partial x(\xi,\theta)}$ and $\frac{\partial t}{\partial x(\xi,\theta)}$? 
 A: What you did is right, but it requires you to obtain $\partial\xi/\partial t$ and $\partial\xi/\partial x$ given $x(\xi,\theta)$ – it would probably be preferable to use the chain rule the other way around to express $u_t$ and $u_x$ in terms of $u_\xi$ and $u_\theta$; that would involve $x_\xi$ and $x_\theta$ instead $\xi_t$ and $\xi_x$; and then you can solve for $u_t$ and $u_x$.
A: The answer is as Joriki stated.
With new transformations
$$u(x,t)\Rightarrow u\big(x(\xi,\theta),t(\theta)\big)$$
The new partial derivatives are
$$\frac{\partial u}{\partial \xi}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}\qquad (1)$$
$$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial t}\frac{dt}{d\theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial t}\qquad (2)$$
From $(1)$ it follows that
$$\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}=\frac{\partial u}{\partial x}\qquad (3)$$
If we substitute $(3)$ into $(2)$
$$\frac{\partial u}{\partial \theta}=\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial t}\Rightarrow \frac{\partial u}{\partial t}=\frac{\partial u}{\partial \theta}-\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}\frac{\partial x}{\partial \theta}\qquad (4)$$
And the original equation becomes
$${\frac{\partial u}{\partial x}}={\frac{\partial u}{\partial t}}\Rightarrow \frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}=\frac{\partial u}{\partial \theta}-\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}\frac{\partial x}{\partial \theta}$$
$$\frac{\partial u}{\partial \xi}=\frac{\partial u}{\partial \theta}\frac{\partial x}{\partial \xi}-\frac{\partial u}{\partial \xi}\frac{\partial x}{\partial \theta}$$
Edit----------
Coordinate transformation is given by
$$x=x(\xi,\theta)$$ $$t=t(\theta)=\theta$$
the the Jacobian can be defined as
$$J(\xi,\theta)=\begin{pmatrix}
\frac{\partial x}{\partial \xi}   & \frac{\partial x}{\partial \theta} \\
\frac{\partial t}{\partial \xi}   & \frac{\partial t}{\partial \theta}
\end{pmatrix}=\begin{pmatrix}
\frac{\partial x}{\partial \xi}   & \frac{\partial x}{\partial \theta} \\
0   & 1
\end{pmatrix}$$
